A straight road generally has the center somewhat higher than the edges to allow water to run off. This is called cross slope or camber. When it is desired to have a banked turn, then the outer edge of the road is raised to produce superelevation, with the outer edge rising above both the center and the inner edge. The bank angle is chosen based on the radius of curvature of the turn and the expected speed of cars going around the turn, while still allowing for the fact that cars might be moving slowly or even stopped. The angle should thus not be chosen to eliminate all friction forces when cars are traveling at maximum speed, as this would be dangerous if traffic had to stop on the road. Velodromes are arenas with tracks designed for high-speed bicycle races, as shown below, with speeds up to 85 km/h. The bank angle on velodrome tracks is chosen to minimize sideways forces on the bicycles when they are traveling at near maximum speeds, so the angle chosen depends on the radius of curvature of the track corners. For example, the Blaine velodrome track pictured below is 250 m long and has a 43° bank angle on the corners and 15° banking on the straightaways. High-speed trains such as the French TGV operate at speeds of over 300 km/h and have run at up to 575 km/h. To accommodate cornering at such speeds, track bends are constructed with a large radius of curvature (at least 7 km for new tracks) and a bank angle of up to about 7° (180 mm maximum superelevation with Standard gauge of 1435 mm). For railways, banking the track is know as cant. An alternative approach for cornering with trains is to leave the track relatively flat and to tilt the train as it travels around a corner. This allows high-speed trains to operate on regular tracks, while maintaining safety and comfort for the passengers. For example, the Queensland Rail Tilt Train operates at 180 km/h by tilting at up to 5° around corners. As we see from our rigid body analysis above, tilting the train will help with avoiding tipping over at high speeds, but will not help with reducing horizontal friction forces. Even with a tilting train, entering a curve at too high a speed will lead to disastrous results.
A car is turning on a banked curve. The coefficient static friction between the car and the pavement is 0.30 and the coefficient of kinetic friction is 0.25.
The angle of the banking is 25 degrees, and the radius of the curve is 50 m. What is the minimum speed the car can have before sliding down the banking. I have found the maximum speed the car can have without sliding up the banking. For the sliding up the bank I did this: b=angle
s= coefficient of static friction Fx = n*sin (b) + sin (b)*n*s = ma Fy = n*cos (b) - mg - cos(b)*n*s = 0 So this is when the car is in equlibrium. From this I find the acceleration and then the speed I haven't figured out a relation between the speed and movement down the banking. I tried to resolve the weight vector into components together with the friction but I didn't get anywhere with that. The weight has some part in this I know. I'm just really lost. Could someone please give me a hint to this problem. The coefficient of kinetic friction is given but I don't see I have to use it. With thanks,
Swatch Last edited: Jun 16, 2005 Answers and Replies
max frictional force can be :: Last edited: Jun 15, 2005
OlderDan
Swatch
Draw a diagram showing the forces acting on the car including weight, normal force, and frictional force. Assume the frictional force is maximum (because you are looking for the minimum speed) and write the friction force in terms of the normal force and the coefficient of friction. Resolve the three forces into x and y components, and try writing your equations again.
I know my first post wasn't to clear. The work I displayed was for the question "What is the maximum speed before the car starts to slide up the banking" in that case the frictional force is pointed down the slope and I get only sine in Fx and cos in Fy. In the case of the question "What is the minimum speed" I did the work again as you asked me to do OlderDan and I succesfully got the right answer. Then I got sine and cos in Fx.
Thanks.
OlderDan
Thanks.
Thanks OlderDan. Of course you're right. I got the angle all mixed up. But the funny thig is I got an answer that was pretty close to the right one. So I made the assumption that I was right, makes you wonder how many times you could be wrong. Thanks for the help.
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What is the maximum speed the automobile can have before sliding up the banking
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